Math for Assembly

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  1. Using Number set 39, 78, 321, 145 convert into ...
  1. Hexadecimal Representation
391639÷16 remainder 72÷16 remainder 2Note:161×2+160×732+7=39∴39 in hexadecimal form is 27781678÷16 remainder 14 or E4÷16 remainder 4Note:161×4+160×1464+14=78∴78 in hexadecimal form is 4E32116321÷16 remainder 120÷16 remainder 41÷16 remainder 1Note:162×1+161×4+160×1256+64+1=321∴321 in hexadecimal form is 14114516145÷16 remainder 19÷16 remainder 9Note:161×9+160×1144+1=145∴145 in hexadecimal form is 91 \begin{aligned} \mathbf{\textcolor{skyblue}{39}} \qquad 16&\sqrt{39} & \div 16\ remainder\ 7\\ &2 & \div 16\ remainder\ 2\\ Note: &\quad 16^1\times 2 + 16^0 \times 7 \\ &\quad 32 + 7 = 39\\ \therefore &\quad\text{\textcolor{khaki}{39 in hexadecimal form is 27}}\\[2em] \mathbf{\textcolor{skyblue}{78}}\qquad 16&\sqrt{78} & \div16\ remainder\ 14\ or\ E\\ &4 & \div16\ remainder\ 4\\ Note: &\quad 16^1 \times 4 + 16^0 \times 14 \\ &\quad 64 + 14 = 78\\ \therefore &\quad\text{\textcolor{khaki}{78 in hexadecimal form is 4E}}\\[2em] \mathbf{\textcolor{skyblue}{321}}\qquad 16&\sqrt{321} & \div16\ remainder\ 1\\ &20 & \div16\ remainder\ 4\\ &1 & \div16\ remainder\ 1\\ Note: &\quad 16^2 \times 1 + 16^1 \times 4 + 16^0 \times 1 \\ &\quad 256 + 64 + 1 = 321\\ \therefore &\quad\text{\textcolor{khaki}{321 in hexadecimal form is 141}}\\[2em] \mathbf{\textcolor{skyblue}{145}}\qquad 16&\sqrt{145} & \div 16\ remainder\ 1\\ &9 & \div 16\ remainder\ 9\\ Note: &\quad 16^1\times 9 + 16^0 \times 1 \\ &\quad 144 + 1 = 145\\ \therefore &\quad\text{\textcolor{khaki}{145 in hexadecimal form is 91}} \end{aligned}
  2. Binary Representation
39239÷2 remainder 119÷2 remainder 19÷2 remainder 14÷2 remainder 02÷2 remainder 01÷2 remainder 1Note:25+22+21+2032+4+2+1=39∴39 in binary form is 10 011178278÷2 remainder 0Same as before question39÷2 remainder 119÷2 remainder 19÷2 remainder 14÷2 remainder 02÷2 remainder 01÷2 remainder 1Note:25+22+21+2032+4+2+1=78∴78 in binary form is 100 11103212321÷2 remainder 1160÷2 remainder 080÷2 remainder 040÷2 remainder 020÷2 remainder 010÷2 remainder 05÷2 remainder 12÷2 remainder 01÷2 remainder 1Note:28+26+20256+64+1=321∴321 in binary form is 1 0100 00011452145÷2 remainder 172÷2 remainder 036÷2 remainder 018÷2 remainder 09÷2 remainder 14÷2 remainder 02÷2 remainder 01÷2 remainder 1Note:27+24+20128+16+1=145∴145 in binary form is 1001 0001 \begin{aligned} \mathbf{\textcolor{skyblue}{39}} \qquad 2&\sqrt{39} & \div2\ remainder\ 1\\ &19 & \div2\ remainder\ 1\\ &9 & \div2\ remainder\ 1\\ &4 & \div2\ remainder\ 0\\ &2 & \div2\ remainder\ 0\\ &1 & \div2\ remainder\ 1\\ Note: &\quad 2^5 + 2^2 + 2^1 + 2^0 \\ &\quad 32 + 4 + 2 + 1 = 39\\ \therefore &\quad\text{\textcolor{khaki}{39 in binary form is 10 0111}}\\[2em] \mathbf{\textcolor{skyblue}{78}}\qquad 2&\sqrt{78} & \div2\ remainder\ 0\\ &\textcolor{orange}{\textbf{Same as before question}}\\ &39 & \div2\ remainder\ 1\\ &19 & \div2\ remainder\ 1\\ &9 & \div2\ remainder\ 1\\ &4 & \div2\ remainder\ 0\\ &2 & \div2\ remainder\ 0\\ &1 & \div2\ remainder\ 1\\ Note: &\quad 2^5 + 2^2 + 2^1 + 2^0 \\ &\quad 32 + 4 + 2 + 1 = 78\\ \therefore &\quad\text{\textcolor{khaki}{78 in binary form is 100 1110}}\\[2em] \mathbf{\textcolor{skyblue}{321}}\qquad 2&\sqrt{321} & \div2\ remainder\ 1\\ &160 & \div2\ remainder\ 0\\ &80 & \div2\ remainder\ 0\\ &40 & \div2\ remainder\ 0\\ &20 & \div2\ remainder\ 0\\ &10 & \div2\ remainder\ 0\\ &5 & \div2\ remainder\ 1\\ &2 & \div2\ remainder\ 0\\ &1 & \div2\ remainder\ 1\\ Note: &\quad 2^8 + 2^6 + 2^0 \\ &\quad 256 + 64 + 1 = 321\\ \therefore &\quad\text{\textcolor{khaki}{321 in binary form is 1 0100 0001}}\\[2em] \mathbf{\textcolor{skyblue}{145}}\qquad 2&\sqrt{145} & \div2\ remainder\ 1\\ &72 & \div2\ remainder\ 0\\ &36 & \div2\ remainder\ 0\\ &18 & \div2\ remainder\ 0\\ &9 & \div2\ remainder\ 1\\ &4 & \div2\ remainder\ 0\\ &2 & \div2\ remainder\ 0\\ &1 & \div2\ remainder\ 1\\ Note: &\quad 2^7 + 2^4 + 2^0 \\ &\quad 128 + 16 + 1 = 145\\ \therefore &\quad\text{\textcolor{khaki}{145 in binary form is 1001 0001}} \end{aligned}

  1. Using Number Set #2 (four binary numbers) 11101101b, 00011011b, 11110101b, 01100100b, show the value of each number as:

    1. An unsigned decimal value
    2. A signed decimal value
272625242322212012864321684211110 1101 b111011012726252423222120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in unsigned decimal form is 23700010011272625242322212016 + 2 + 1 = 19∴1110 1101 in signed decimal form is -190001 1011 b00011011272625242322212016 + 8 + 2 + 1 = 27∴0001 1011 in unsigned decimal form is 27111011012726252423222120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in signed decimal form is -2371111 0101 b111101012726252423222120128 + 64 + 32 + 16 + 4 + 1 = 245∴1111 0101 in unsigned decimal form is 2450000101127262524232221208 + 2 + 1 = 11∴1111 0101 in signed decimal form is -110110 0100 b01100100272625242322212064 + 32 + 4 = 100∴0110 0100 in unsigned decimal form is 100100110112726252423222120128 + 16 + 8 + 2 + 1 = 155∴0110 0100in signed decimal form is -155 \begin{matrix*}[c] 2^7 & 2^6 & 2^5 & 2^4 & 2^3 & 2^2 & 2^1 & 2^0 &\\ 128 & 64 & 32 & 16 & 8 & 4 & 2 & 1 &\\\\ \end{matrix*}\\[1em] \textbf{\textcolor{skyblue}{1110 1101 b}}\\[2em] \begin{matrix*}[c] 1&1&1&0& 1&1&0&1&\\ 2^7 & 2^6 & 2^5 & \phantom{ 2^4 } & 2^3 & 2^2 & \phantom{ 2^1 } & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 64 + 32 + 8 + 4 + 1 = 237} \\ \therefore \quad\text{\textcolor{khaki}{1110 1101 in unsigned decimal form is 237}}\\[1em] \begin{matrix*}[c] 0&0&0&1& 0&0&1&1&\\ \phantom{ 2^7 } & \phantom{ 2^6 } & \phantom{ 2^5 } & 2^4 & \phantom{ 2^3 } & \phantom{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{16 + 2 + 1 = 19} \\ \therefore \quad\text{\textcolor{khaki}{1110 1101 in signed decimal form is -19}}\\[2em] \textbf{\textcolor{skyblue}{0001 1011 b}}\\[2em] \begin{matrix*}[c] 0&0&0&1& 1&0&1&1&\\ \phantom{ 2^7 } & \phantom{ 2^6 } & \phantom{ 2^5 } & 2^4 & 2^3 & \phantom{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{16 + 8 + 2 + 1 = 27} \\ \therefore \quad\text{\textcolor{khaki}{0001 1011 in unsigned decimal form is 27}}\\[2em] \begin{matrix*}[c] 1&1&1&0& 1&1&0&1&\\ 2^7 & 2^6 & 2^5 & \phantom{ 2^4 } & 2^3 & 2^2 & \phantom{ 2^1 } & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 64 + 32 + 8 + 4 + 1 = 237} \\ \therefore \quad\text{\textcolor{khaki}{1110 1101 in signed decimal form is -237}}\\[2em] \textbf{\textcolor{skyblue}{1111 0101 b}}\\[2em] \begin{matrix*}[c] 1&1&1&1& 0&1&0&1&\\ 2^7 & 2^6 & 2^5 & 2^4 & \phantom{ 2^3 } & 2^2 & \phantom{ 2^1 } & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 64 + 32 + 16 + 4 + 1 = 245} \\ \therefore \quad\text{\textcolor{khaki}{1111 0101 in unsigned decimal form is 245}}\\[2em] \begin{matrix*}[c] 0&0&0&0& 1&0&1&1&\\ \phantom{}{ 2^7 } & \phantom{}{ 2^6 } & \phantom{}{ 2^5 } & \phantom{}{ 2^4 } & 2^3 & \phantom{}{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{8 + 2 + 1 = 11} \\ \therefore \quad\text{\textcolor{khaki}{1111 0101 in signed decimal form is -11}}\\[2em] \textbf{\textcolor{skyblue}{0110 0100 b}}\\[2em] \begin{matrix*}[c] 0&1&1&0& 0&1&0&0&\\ \phantom{ 2^7 } & 2^6 & 2^5 & \phantom{ 2^4 } & \phantom{ 2^3 } & 2^2 & \phantom{ 2^1 } & \phantom{ 2^0 } &\\ \end{matrix*}\\[1em] \textsf{64 + 32 + 4 = 100} \\ \therefore \quad\text{\textcolor{khaki}{0110 0100 in unsigned decimal form is 100}}\\[2em] \begin{matrix*}[c] 1&0&0&1& 1&0&1&1&\\ 2^7 & \phantom{ 2^6 } & \phantom{ 2^5 } & 2^4 & 2^3 & \phantom{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 16 + 8 + 2 + 1 = 155} \\ \therefore \quad\text{\textcolor{khaki}{0110 0100in signed decimal form is -155}}\\[2em]
  1. [10] Using Number Set #2 (four binary numbers) 01100100b, 10011010b, 01101101b, 11000110b, where the first number is considered to be A, the second number B, the third C, etc… calculate each of the following:
  1. A v B, A v C, A v D ( v is the symbol for the OR operation) 
    </li>
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    A ^ B, A ^ C, A ⊕ D (^ is the symbol for the AND operation, ⊕ is the symbol for the XOR operation)

    </li>
  1. Using Number Set #2 (four binary numbers) 11101101b, 00011011b, 11110101b, 01100100b, show the value of each number as:
  1. An unsigned decimal value 
    </li>
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  A signed decimal value
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272625242322212012864321684211110 1101 b111011012726252423222120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in unsigned decimal form is 23700010011272625242322212016 + 2 + 1 = 19∴1110 1101 in signed decimal form is -190001 1011 b00011011272625242322212016 + 8 + 2 + 1 = 27∴0001 1011 in unsigned decimal form is 27111011012726252423222120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in signed decimal form is -2371111 0101 b111101012726252423222120128 + 64 + 32 + 16 + 4 + 1 = 245∴1111 0101 in unsigned decimal form is 2450000101127262524232221208 + 2 + 1 = 11∴1111 0101 in signed decimal form is -110110 0100 b01100100272625242322212064 + 32 + 4 = 100∴0110 0100 in unsigned decimal form is 100100110112726252423222120128 + 16 + 8 + 2 + 1 = 155∴0110 0100in signed decimal form is -155 \begin{matrix*}[c] 2^7 & 2^6 & 2^5 & 2^4 & 2^3 & 2^2 & 2^1 & 2^0 &\\ 128 & 64 & 32 & 16 & 8 & 4 & 2 & 1 &\\\\ \end{matrix*}\\[1em] \textbf{\textcolor{skyblue}{1110 1101 b}}\\[2em] \begin{matrix*}[c] 1&1&1&0& 1&1&0&1&\\ 2^7 & 2^6 & 2^5 & \phantom{ 2^4 } & 2^3 & 2^2 & \phantom{ 2^1 } & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 64 + 32 + 8 + 4 + 1 = 237} \\ \therefore \quad\text{\textcolor{khaki}{1110 1101 in unsigned decimal form is 237}}\\[1em] \begin{matrix*}[c] 0&0&0&1& 0&0&1&1&\\ \phantom{ 2^7 } & \phantom{ 2^6 } & \phantom{ 2^5 } & 2^4 & \phantom{ 2^3 } & \phantom{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{16 + 2 + 1 = 19} \\ \therefore \quad\text{\textcolor{khaki}{1110 1101 in signed decimal form is -19}}\\[2em] \textbf{\textcolor{skyblue}{0001 1011 b}}\\[2em] \begin{matrix*}[c] 0&0&0&1& 1&0&1&1&\\ \phantom{ 2^7 } & \phantom{ 2^6 } & \phantom{ 2^5 } & 2^4 & 2^3 & \phantom{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{16 + 8 + 2 + 1 = 27} \\ \therefore \quad\text{\textcolor{khaki}{0001 1011 in unsigned decimal form is 27}}\\[2em] \begin{matrix*}[c] 1&1&1&0& 1&1&0&1&\\ 2^7 & 2^6 & 2^5 & \phantom{ 2^4 } & 2^3 & 2^2 & \phantom{ 2^1 } & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 64 + 32 + 8 + 4 + 1 = 237} \\ \therefore \quad\text{\textcolor{khaki}{1110 1101 in signed decimal form is -237}}\\[2em] \textbf{\textcolor{skyblue}{1111 0101 b}}\\[2em] \begin{matrix*}[c] 1&1&1&1& 0&1&0&1&\\ 2^7 & 2^6 & 2^5 & 2^4 & \phantom{ 2^3 } & 2^2 & \phantom{ 2^1 } & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 64 + 32 + 16 + 4 + 1 = 245} \\ \therefore \quad\text{\textcolor{khaki}{1111 0101 in unsigned decimal form is 245}}\\[2em] \begin{matrix*}[c] 0&0&0&0& 1&0&1&1&\\ \phantom{}{ 2^7 } & \phantom{}{ 2^6 } & \phantom{}{ 2^5 } & \phantom{}{ 2^4 } & 2^3 & \phantom{}{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{8 + 2 + 1 = 11} \\ \therefore \quad\text{\textcolor{khaki}{1111 0101 in signed decimal form is -11}}\\[2em] \textbf{\textcolor{skyblue}{0110 0100 b}}\\[2em] \begin{matrix*}[c] 0&1&1&0& 0&1&0&0&\\ \phantom{ 2^7 } & 2^6 & 2^5 & \phantom{ 2^4 } & \phantom{ 2^3 } & 2^2 & \phantom{ 2^1 } & \phantom{ 2^0 } &\\ \end{matrix*}\\[1em] \textsf{64 + 32 + 4 = 100} \\ \therefore \quad\text{\textcolor{khaki}{0110 0100 in unsigned decimal form is 100}}\\[2em] \begin{matrix*}[c] 1&0&0&1& 1&0&1&1&\\ 2^7 & \phantom{ 2^6 } & \phantom{ 2^5 } & 2^4 & 2^3 & \phantom{ 2^2 } & 2^1 & 2^0 &\\ \end{matrix*}\\[1em] \textsf{128 + 16 + 8 + 2 + 1 = 155} \\ \therefore \quad\text{\textcolor{khaki}{0110 0100in signed decimal form is -155}}\\[2em]

  1. Using Number Set #2 (four binary numbers) 01100100b, 10011010b, 01101101b, 11000110b, where the first number is considered to be A, the second number B, the third C, etc… calculate each of the following:

    1. A v B, A v C, A v D ( v is the symbol for the OR operation)
    2. A ^ B, A ^ C, A ⊕ D (^ is the symbol for the AND operation, ⊕ is the symbol for the XOR operation)
0110 0100A∨B=∨1001 10101111 1110‾0110 0100A∨C=∨0110 11010110 1101‾0110 0100A∨D=∨1100 01101110 0110‾0110 0100A∧B=∧1001 10100000 0000‾0110 0100A∧C=∧0110 11010110 0100‾0110 0100A⊕D=⊕1100 01101010 0010‾ \begin{aligned} && 0110\ 0100 \\ A\lor B =\quad& \lor& 1001\ 1010\\ &&\overline{1111\ 1110} \end{aligned} \\[1em] \begin{aligned} && 0110\ 0100 \\ A\lor C =\quad& \lor& 0110\ 1101\\ &&\overline{0110\ 1101} \end{aligned} \\[1em] \begin{aligned} && 0110\ 0100 \\ A\lor D =\quad& \lor& 1100\ 0110\\ &&\overline{1110\ 0110} \end{aligned} \\[1em] \begin{aligned} && 0110\ 0100 \\ A\wedge B =\quad& \wedge& 1001\ 1010\\ &&\overline{0000\ 0000} \end{aligned} \\[1em] \begin{aligned} && 0110\ 0100 \\ A\wedge C =\quad& \wedge& 0110\ 1101\\ &&\overline{0110\ 0100} \end{aligned} \\[1em] \begin{aligned} && 0110\ 0100 \\ A\oplus D =\quad& \oplus& 1100\ 0110\\ &&\overline{1010\ 0010} \end{aligned}

  1. Using Number Set#3 - HEX 71AF2523h, 2B988398h, 9E5E4AD8h, 6B7C3487h (four hexadecimal numbers) where the first number is Q, the second number R, the third S, etc… calculate each of the following:

    1. Q + R, Q + S, Q + T (show the carry value – 9th digit, if there is a carry value)

      There is a carry for the 9th digit for Q + S\tt \textcolor{khaki}{\text{There is a carry for the 9th digit for Q + S}}

    2. Q – R, Q – S, Q – T (use the 16’s compliment method)
111F 252371AF 2523+ 2B98 83989D47 A8BB‾71AF 2523− 2B98 839816’s complement of R FFFF FFFF− 2B98 8398+ 2B98 8391D467 7C68‾10111 1000 D467 7C68+ 71AF 25234616 A18B‾1111F 252371AF 2523+ 9E5E 4AD8100D 6FFB‾71AF 2523− 9E5E 4AD816’s complement of S FFFF FFFF− 9E5E 4AD8+ 2B98 839161A1 B528‾0111 0000 61A1 B528+ 71AF 2523D350 DA4A‾111F 252371AF 2523+ 6B7C 3487DD2B 59AA‾71AF 2523− 6B7C 348716’s complement of T FFFF FFFF− 6B7C 3487+ 2B98 83919483 CB79‾10111 1000 9483 CB79+ 71AF 25230632 F09C‾ \begin{aligned} \tt \phantom{1} 1 1 \phantom{F\ 2523} &\\ \tt 7 1 A F\ 2523 &\\ +\ \tt 2B98\ 8398 &\\ \tt \textcolor{khaki}{ \overline{9D47\ A8BB}}& \\[1em] \tt 71AF\ 2523 &\\ -\ \tt 2B98\ 8398 &\\ \\ \tiny\text{16's complement of R} \\ \ \tt FFFF\ FFFF &\\ -\ \tt 2B98\ 8398 &\\ +\ \tt \phantom{2B98\ 839}1 &\\ \tt { \overline{D467\ 7C68}}& \\[1em] \tt 1 \phantom0 1 1 \phantom{1\ } 1\phantom{000} &\\ \ \tt D467\ 7C68 &\\ +\ \tt 71AF\ 2523 &\\ \tt \textcolor{khaki}{ \overline{4616\ A18B}}& \\[1em] \end{aligned} \begin{aligned} \tt 1 1 1 1 \phantom{F\ 2523} &\\ \qquad \tt 71AF\ 2523 &\\ \qquad +\ \tt 9E5E\ 4AD8 &\\ \tt \textcolor{khaki}{ \overline{100D\ 6FFB}}& \\[1em] \qquad \tt 71AF\ 2523 &\\ \qquad -\ \tt 9E5E\ 4AD8 &\\ \\ \tiny\text{16's complement of S} \\ \ \tt FFFF\ FFFF &\\ -\ \tt 9E5E\ 4AD8 &\\ +\ \tt \phantom{2B98\ 839}1 &\\ \tt { \overline{61A1\ B528}}& \\[1em] \tt \phantom0 1 1 \phantom{1\ } \phantom{0000} &\\ \ \tt 61A1\ B528 &\\ +\ \tt 71AF\ 2523 &\\ \tt \textcolor{khaki}{ \overline{D350\ DA4A}}& \\[1em] \end{aligned} \begin{aligned} \tt \phantom{1} 1 1 \phantom{F\ 2523} &\\ \qquad \tt 71AF\ 2523 &\\ \qquad +\ \tt 6B7C\ 3487 &\\ \tt \textcolor{khaki}{ \overline{DD2B\ 59AA}}& \\[1em] \qquad \tt 71AF\ 2523 &\\ \qquad -\ \tt 6B7C\ 3487 &\\ \\ \tiny\text{16's complement of T} \\ \ \tt FFFF\ FFFF &\\ -\ \tt 6B7C\ 3487 &\\ +\ \tt \phantom{2B98\ 839}1 &\\ \tt { \overline{9483\ CB79}}& \\[1em] \tt 1 \phantom0 1 1 \phantom{1\ } 1\phantom{000} &\\ \ \tt 9483\ CB79 &\\ +\ \tt 71AF\ 2523 &\\ \tt \textcolor{khaki}{ \overline{0632\ F09C}}& \\[1em] \end{aligned} Q:71AF 2523 hR:2B98 8398 hS:9E5E 4AD8 hT:6B7C 3487 h \mathtt{\textcolor{skyblue}{Q: 71AF\ 2523 \ h}}\\[1em] \mathtt{\textcolor{skyblue}{R: 2B98\ 8398 \ h}}\\[1em] \mathtt{\textcolor{skyblue}{S: 9E5E\ 4AD8 \ h}}\\[1em] \mathtt{\textcolor{skyblue}{T: 6B7C\ 3487 \ h}}\\[1em]

For My Reference

Decimal Hexadecimal Binary
0 0 0000
1 1 0001
2 2 0010
3 3 0011
4 4 0100
5 5 0101
6 6 0110
7 7 0111
8 8 1000
9 9 1001
10 A 1010
11 B 1011
12 C 1100
13 D 1101
14 E 1110
15 F 1111